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Proof step 4 from proof prefix-append-one-sizeeq (previous step)

# x0 = # x1 + 1, ∃ z. x0 + z = x1 + a ' ⊦ x0 = x1 + a '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# x0 = # x1 + 1, x0 + z = x1 + a ', a ∈ z ⊦ z = []

Rule argument

Additional information

Used simplifier rules: ni: ⊦ x + a ' = y + z ∧ ¬ a ∈ z → z = []; (forward) from specification list% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ x = x + y ↔ y = []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list